How to prove that there is no r in the rationals such that r^2 = 6

Question

 Can someone help me write a full proof for this

Answer 1

Assume r=p/q where p and q are integers with gcd(p,q)=1. Then r^2=6=(P^2)/(q^2), or

 

2*3*q*q=p*p.   (1)

 

 

Since 2 divides LHS of (1) then it must also divide RHS of (1) i.e. 2 divides p, so p=2k for some integer k. Then (1) gives us

 

2*3*q*q=2*k*2*k or 3*q*q=2*k*k    (2)

 

But since 2 divides RHS of (2) it must also divide LHS of (1) and therefore it must divide q. But this implies that 2 divides both p and q which is a contradiction to the assumption that gsd(p,q)=1.

 

QED

Answer 2

r² = 6

 

r = √(6)

 

The number 6 is not a perfect square.  Also, the square root of 6 cannot be written as a simple fraction because there many significant non-repeating digits after the decimal point.